Maths

142 MATHEMATICS
We need to prove that
2 2 2
✁ ✁ ✁
✂ ✂ ✂ ✄ ☎ ✆
☎ ✆ ☎ ✆
ar (ABC) AB BC CA
ar (PQR) PQ QR RP
✝ ✞ ✝ ✞
For finding the areas of the two triangles, we draw altitudes AM and PN of the
triangles.
✝
✞
Now,
ar (ABC) = 1 BC × AM
2
and
So,
Now, in ✡ ABM and ✡ PQN,
ar (PQR) = 1 QR × PN
2
1
ar (ABC)
BC × AM
✟
BC × AM
ar (PQR) = ✠
2
1
QR × PN
QR × PN
✟
2
☛ ☛ ✡ ✡
☛ ☛
B = Q
(As ABC ~ PQR)
and M = N (Each is of 90°)
So, ✡ ABM ~ ✡ PQN (AA similarity criterion)
✡ ✡
Therefore,
AM
PN = AB
PQ
(2)
Also, ABC ~ PQR (Given)
So,
(1)
AB
PQ = BC CA ☞ (3)
QR RP
Therefore,
ar (ABC)
ar (PQR) = AB AM ✌ [From (1) and (3)]
PQ PN
=
=
Now using (3), we get
AB AB
✌ [From (2)]
PQ PQ
AB ✁
☎ ✆
✝ ✞ PQ
2 2 2
ar (ABC)
ar (PQR) = ✁ AB ✁ BC ✁ CA
✂ ☎ ✆
☎ ✆ ☎ ✆ ✍
✂
✞
✝ ✞ ✝ ✞
PQ QR RP ✝
Let us take an example to illustrate the use of this theorem.
2